anonymous
  • anonymous
Infinite Series - Find the interval of convergence for the following geometric series and, within this interval, find the sum; \[\sum_{n=2}^{Infinity} (x-1)^{n-1}/2^{n+1}\] \[\sum_{n=0}^{Infinity} (x-1)^{n+1}/2^{n+3}\] =(x-1)^n (x-1)^1 / (2^n * 2^3) = (x-1)/8 * (x-1/2)^n Sort of stuck where to go from here. I wasn't sure if A = (x-1)/8 and r= (x-1)/2 and I just use a/(1-r) or what. Any help would be appreciated!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
are you trying to find where the 2 of them meet?
anonymous
  • anonymous
Not entirely sure what that means.
anonymous
  • anonymous
The sandwich method? I was using that for Sequences.

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amistre64
  • amistre64
well, ive done intervals of convergence on a single summation; but ive never even considered doing it for 2 at the same time
amistre64
  • amistre64
i assume it wants the interval where they meet at
anonymous
  • anonymous
Again, not sure what you mean by 2 at the same time. If you mean the first two equations, the first is N=2 and then I did N = 0 and added two to N.
anonymous
  • anonymous
If you have another method in mind for solving it I would love to hear it. I think what is getting me is the "X". If it were a 2, instead of X, I could have 1/8 instead of (x-1)/8, which would be my A, and then (x-1)/2 would be my R and I would use a/(1-r)
amistre64
  • amistre64
\[\lim_{n\to inf}\frac{(x-1)^{n-1}}{2^{n+1}}\frac{2^{n}}{(x-1)^{n-2}}\] \[\lim_{n\to inf}\frac{(x-1)}{2}=\frac{x-1}{2}\]
amistre64
  • amistre64
i thought you had to sums there to play with; but it looks like you just altered the second one from teh first
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
anonymous
  • anonymous
Yeah - random question - how do you get the division to go one on top of the other?
anonymous
  • anonymous
Thanks for the link, Ill check it out.
amistre64
  • amistre64
the limit is a ratio of \[\frac{a_n}{a_{n-1}}\] and the fraction thing is the code: \frac{top}{bottom}
amistre64
  • amistre64
if im reading the site right; the interval of convergence is then: \[\frac{|x-1|}{2}<1\] \[|x-1|<2\] -1 < x < 3
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sum+%28x%E2%88%921%29%5E%28n%E2%88%921%29+%2F%282%5E%28n%2B1%29%29+from+0+to+inf++ and i think the wolf agrees ;)

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